Will Cadmium hydroxide precipitate from 0.01 M solution of #"CdCl"_2# at pH 9. #K_(sp)# of #"Cd"("OH")_2 = 2.5 xx 10^(-14)"M"^3#? Please solve,thanks!
1 Answer
Yes, a precipitate will form.
Explanation:
Your strategy here will be to calculate the concentrations of the cadmium(II) cations,
So, you know that the pH of the solution is equal to
#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " POH" = 14)color(white)(a/a)|)))#
This means that the pOH of the solution, which can be used to find the concentration of hydroxide anions, will be equal to
#"pOH" = 14 - 9 = 5#
This means that you have
#["OH"^(-)] = 10^(-"pOH") = 10^(-5)"M"#
Cadmium chloride,
#"CdCl"_text(2(aq]) -> "Cd"_text((aq])^(2+) |+ 2"Cl"_text((aq])^(-)#
Notice that one mole of cadmium chloride produces one mole of cadmium cations in solution. This means that you have
#["Cd"^(2+)] = ["CdCl"_2] = 10^(-2)"M"#
The partial dissociation of cadmium hydroxide in aqueous solution looks like this
#"Cd"("OH")_text(2(s]) rightleftharpoons "Cd"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#
By definition, the solubility product constant,
#K_(sp) = ["Cd"^(2+)] * ["OH"^(-)]^color(red)(2)#
Now, in order to determine if a precipitate if formed or not, you need to calculate the reaction quotient,
The difference between
In other words,
#Q_(sp) > K_(sp)#
At this point, a precipitate will form until you reach saturation, i.e until
If
Plug in the concentrations of the two ions to get
#Q_(sp) = 10^(-2)"M" * (10^(-5))^color(red)(2)"M"^color(red)(2) = 10^(-12)"M"^3#
Since you have
#10^(-12)"M"^3 > 2.5 * 10^(-14)"M"^3#
cadmium hydroxide will indeed precipitate out of solution.